Now suppose for each $i$ we have injective homomorphisms from $F_i$ to $K_{\sigma(i)}$ and from $K_i$ to $F_{\mu(i)}$ where $i \leq \sigma(i)$ and $i \leq \mu(i)$. In other words, each field $F_i$ is isomorphic to a subfield of some $K_j$ where $j \geq i$ and each field $K_i$ is isomorphic to a subfield of some $F_j$ where $j \geq i$. [Think of the two towers sitting next to each other with arrows pointing diagonally upward.]

My question, can we conclude that $F \cong K$?

A colleague asked me this question some time ago. I came up with a sketch of a proof for the case when $F_{i+1}$ is an algebraic extension of $F_i$ and $K_{i+1}$ is an algebraic extension of $K_i$ for each $i$. I suspect it's false in general [something to do with the fact that injective and surjective aren't equivalent for maps between infinite dimensional spaces.]

Does anybody know a counterexample for the general case?

I would also appreciate a reference for the algebraic case (where I'm 99% sure it's true).

The answers below give very nice examples where this is false, but the result is still true in some common cases. For example, the result is true when all the F's and K's are number fields. More generally, I think this is true if the F's and K's are all finite extensions of some base field and the embeddings are required to fix that base field. Are there broader circumstances where the result is still true?
–
François G. Dorais♦Sep 6 '11 at 11:15

Many Thanks Francois! I should have been more careful when I posed the question. After @JSpecter posted his counterexample, I started to think back about my "proof" and realized I had been working in the context where $F_0=K_0=\mathbb{Q}$ (and all extensions are algebraic). I too would like to know when something like this holds (and get some references).
–
Bill CookSep 7 '11 at 17:51

2 Answers
2

This is false even in the case you describe as it would imply Cantor-Schroeder-Bernstein for fields. That said, we can take the standard counter example for the claim of CSB for fields as a counter example for your claim. Let $F_i = \mathbb{C}(X)$ and $K_i = \mathbb{C}$ for all $i.$ Then $K_i$ injects into $F_i$ and $F_i$ injects into $K_i$ for all $i,$ but the unions $F$ and $K$ are not isomorphic, as $F$ is not algebraically closed.

Thanks! I guess my "sketch" of a "proof" was flawed. What is this injection from rational functions into the complex numbers? I guess I'm not familiar with that example.
–
Bill CookSep 6 '11 at 3:22

1

Both $\mathbb{C}(X)$ and $\mathbb{C}$ have the same transcendence degree over $\mathbb{Q},$ so there exists an isomorphism between their algebraic closures. Restricting this isomorphism to $\mathbb{C}(X)$ gives the desired injection.
–
JSpecterSep 6 '11 at 3:38

Here is a counterexample which, unlike JSpecter's, does not rely on transcendence degrees and Zorn's lemma (used to create an embedding of ${\mathbf C}(X)$ into $\mathbf C$). It comes from a pair of isogeneous but nonisomorphic elliptic curves. Let $E_1$ and $E_2$ be elliptic curves over ${\mathbf Q}$ admitting an isogeny $E_1 \rightarrow E_2$. There is a (dual) isogeny $E_2 \rightarrow E_1$ and these maps provide one with homomorphisms between the function fields ${\mathbf Q}(E_2) \rightarrow {\mathbf Q}(E_1)$ and ${\mathbf Q}(E_1) \rightarrow {\mathbf Q}(E_2)$. So each function field embeds into the other. Choosing $E_1$ and $E_2$ to be isogeneous but not isomorphic over ${\mathbf Q}$ -- say their $j$-invariants are not equal to assure non-isomorphism -- the function fields ${\mathbf Q}(E_1)$ and ${\mathbf Q}(E_2)$ are not isomorphic as fields. This is an absolute statement since we work over ${\mathbf Q}$ rather than, say, ${\mathbf C}$ (where one would conclude the function fields are not isomorphic over ${\mathbf C}$, i.e., fixing ${\mathbf C}$). The same argument goes through using isogenous but non-isomorphic elliptic curves over ${\mathbf F}_p$ for any prime $p$.